Question 1099282
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Factor the Polynomial completely

p(x)= x^5+6x^3+9x
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<pre>
{{{x^5+6x^3+9x}}} = {{{x*(x^4 + 6x^2 + 9)}}} =    


     ! Notice that x^4 + 6x^2 + 9 = {{{(x^2)^2 + 2*3*x^2 + 9}}} = {{{(x^2+3)^2}}}.  
       Therefore, you can continue this chain of equalities in this way


= {{{x*(x^2+3)^2}}} =


      The only real zero is z= 0.  All other zeroes are complex zeroes.
      So, you can continue in the complex domain


= {{{x*(x-3i)^2*(x+3i)^2}}}.


      Now you can see that there is one real zero x= 0 of multiplicity 1,

                                        complex zero x= 3i of multiplicity 2, and

                                        complex zero  x= -3i of multiplicity 2.
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